# Answer to Question #25252 in Abstract Algebra for Mohammad

Question #25252

Show that from: rad (R[t]) = (Nil*R)[t] follows Kothe’s Conjecture. (“The sum of two nil left ideals in any ring is nil”.)

Expert's answer

We shall prove Kothe’s Conjecture inthe alternative form: If

since

*R*is a ring such that Nil**R*= 0, thenevery nil left ideal*I**⊆**R*is zero. Itsuffices to show that, for any*f*(*t*)*∈**I*[*t*], 1 +*f*(*t*) has a leftinverse in*R*[*t*], for then*I*[*t*]*⊆**rad**R*[*t*], and rad*R*[*t*] = 0 by assumption.Consider the ring*R'*=*I**⊕**Z, obtained by formally adjoining anidentity 1**'*= (0*,*1) to the “rng”*I*. It is easy to seethat Nil**R'*=*I*, so by assumption, rad*R'*[*t*] =*I*[*t*].Given*f*(*t*)*∈**I*[*t*],1*'*+*f*(*t*) has then a left inverse, say 1*'*+*g*(*t*),in*R'*[*t*]. From1*'*= (1*'*+*g*(*t*))(1*'*+*f*(*t*))*,*we have*−g*(*t*) =*f*(*t*) +*g*(*t*)*f*(*t*)*∈**I*[*t*]*,*since

*I*[*t*] is an idealin*R'*[*t*]. Now the equation 1 = (1+*g*(*t*))(1 +*f*(*t*))implies that 1 +*f*(*t*) has a left inverse 1 +*g*(*t*)*∈**R*[*t*], as desired.
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